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bishop1989
bishop1989
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I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete CAT$(0)$$CAT(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isomorphicisometric embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete CAT$(0)$$CAT(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for CAT$(0)$ $CAT(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumptionsassumsions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for CAT$(0)$$CAT(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.

I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete CAT$(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isomorphic embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete CAT$(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for CAT$(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumptions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for CAT$(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.

I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete $CAT(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isometric embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete $CAT(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for $CAT(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumsions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for $CAT(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.
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YCor
YCor
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Amalgamated Product Acting ofproduct acting on CAT(0) Cube Complexcube complex

I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete $CAT(0)$CAT$(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isomorphic embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete $CAT(0)$CAT$(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for $CAT(0)$CAT$(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumsionsassumptions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for $CAT(0)$CAT$(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.

Amalgamated Product Acting of CAT(0) Cube Complex

I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete $CAT(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isomorphic embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete $CAT(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for $CAT(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumsions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for $CAT(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.

Amalgamated product acting on CAT(0) cube complex

I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete CAT$(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isomorphic embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete CAT$(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for CAT$(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumptions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for CAT$(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.
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bishop1989
bishop1989
  • 349
  • 1
  • 3

Amalgamated Product Acting of CAT(0) Cube Complex

I was reading the following result from the book Metric spaces of non-positive curvature by Bridson and Haefliger.

Result: Let $F_0,F_1$ and $H$ be groups acting properly by isometries on complete $CAT(0)$ spaces $X_0, X_1$ and $Y$ respectively. Suppose that for $j=0,1 $ there exists a monomorphism $\phi_j:H \to F_j$ and a $\phi_j$-equivariant isomorphic embedding $f_j:Y\to X_j$. Then

(1) the amalgamated product $G=F_0\ast_H F_1$ associated to the maps $\phi_j$ acts properly by isometries on a complete $CAT(0)$ space $X$;

(2) if the given action of $F_0$, $F_1$ and $H$ are cocompact, then the action of $G$ on $X$ is cocompact.

I'd be happy to see proof of the above result for $CAT(0)$ (finite dimensional if needed) cube complexes with the same hypothesis as above. If it is false what may go wrong?

From a quick google search I found the paper "Cubulating malnormal amalgams" by Hsu and Wise (https://link.springer.com/article/10.1007/s00222-014-0513-4) and saw a similar result (see Theorem A) but with extra assumsions like resulting group is hyperbolic relative to free abelian and the edge group hyperbolic, quasi-convex, malnormal etc.

What may go wrong if we simply try to extend the above result for $CAT(0)$ cube complexes without assuming these conditions of Theorem A (by Hsu & Wise)?

I think the edge subgroup must act on a convex subset (that sits inside both the bigger cube complexes) such a way that hyperplanes are "merged" in both the other spaces while patching or some sort of wall compatibility condition must be satisfied.